Store A uses the newsvendor model to manage its inventory. Demand for its product is normally distributed with a mean of 500 and a standard deviation of 300. What is its in-stock probability if Store A’s order quantity is 800 units?

Answer:[tex]P(X<800)=P(Z<1)=0.841[/tex]Step-by-step explanation:Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  Let X the random variable that represent the Demand for its product on this case, and for this case we know the distribution for X is given by:  [tex]X \sim N(\mu=500,\sigma=300)[/tex]  And let [tex]\bar X[/tex] represent the sample mean, the distribution for the sample mean is given by:  [tex]\bar X \sim N(\mu,\frac{\sigma}{\sqrt{n}})[/tex] What is its in-stock probability if Store A’s order quantity is 800 units?We are looking for this probability:What is its in-stock probability if Store A’s order quantity is 800 units?So we can find the following values:[tex]P(X>800)[/tex] and [tex]P(X<800)[/tex]Sor this problem we can use the z score formula given by:[tex]z=\frac{x-\mu}{\sigma}[/tex]If we find the z score for the value 800 we got:[tex]z=\frac{800-500}{300}=1[/tex]And if we find:[tex]P(X<800)=P(Z<1)=0.841[/tex]And by the complement rule:[tex]P(X>800)=1-P(X<800)= 1-0.841=0.159[/tex]